How much tribromaniline will be obtained from 340 g of aniline c6h5nh2 when it interacts with Br2?

Let’s execute the solution:

According to the condition of the problem, we write the data:

C6H5NH2 + 3Br2 = C6H2NH2Br3 + 3HBr – substitution, tribromaniline was formed;

Calculations:
M (C6H5NH2) = 93 g / mol;

M (tribromoaniline) = 329.9 g / mol;

Y (C6H5NH2) = m / M = 340/93 = 3.7 mol;

Y (tribromaniline) = 3.7 mol since the amount of substances is 1 mol.

Find the mass of the product:
m (tribromaniline) = Y * M = 3.7 * 329.9 = 1220.6 g = 1 kg 220 g

Answer: Tribromaniline weighing 1 kg 220 g was obtained